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My guess will be

$\Delta S_\mathrm{(system)}$ : increase [because heat is flowing into the system]

$\Delta S_\mathrm{(surrounding)}$: decrease [because heat is leaving the environment]

$\Delta S_\mathrm{(universe (sys+surr))}$ : $0$, because it is a reversible process.

Please confirm if my reasonings are correct. Thank you

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  • $\begingroup$ by definition reversibly does not change entropy (remember entropy difference can be zero as well) $\endgroup$ – Nikos M. Sep 30 '14 at 2:38
  • $\begingroup$ what do you mean by constant T in the title ? you must edit the title ( it is : "... is transfered in a reversible process ?" ? ) $\endgroup$ – user46925 Dec 26 '15 at 22:57
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Yes you are right. Entropy of universe or isolated system is always generated (for irreversible process) or equal to zero ( for reversible process)

Entropy of system increase because of heat addition and that of surrounding decreases $\left(-{\frac{\Delta Q}{T}}\right)$.

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